Ring of invariants for $n$-tuples of Lie algebras $\DeclareMathOperator\GL{GL}\DeclareMathOperator\M{M}\DeclareMathOperator\Tr{Tr}$Consider the diagonal action of $\GL(n,\mathbb{C})$ on the variety of $k$-tuples of matrices, $\M_{n\times n}(\mathbb{C})^k$
through conjugation.  Then it is known that the ring of invariants
$\mathbb{C}[\M_{n\times n}(\mathbb{C})^k]^{\GL(n,\mathbb{C})}$
is generated by the functions obtained by first evaluating a non-commutative polynomial on the tuple of matrices and then applying the trace to the resulting matrix.
So for example if $k=2$, we need to look at functions like $(A,B) \mapsto \Tr(AB)$, $(A,B) \mapsto \Tr((AB)^2 A)$, etc.  Details can be found in:
The invariant theory of $n \times n$ matrices, Claudio Procesi, Advances in Mathematics, Volume 19, Issue 3, March 1976, Pages 306-381
Now assuming this, suppose we are looking for the ring of invariants of tuples as before but now for the diagonal action of a reductive group $G$ on the variety of tuples $\mathfrak{g}^k$.
Then will it be true that the ring of invariants in this situation is also generated as above by first evaluating non-commutative polynomials and then taking the trace, but now we do this on the matrices obtained by considering various representations of $\mathfrak{g}$ (or $G$), just like what happens in the case of $k=1$?
 A: The orthogonal and symplectic  case are treated in my old paper as well, there is a paper by Gerry Schwarz  on G_2,  always in characteristic 0, the theory in positive characteristic is presented  in
The Invariant Theory of Matrices  – 30 dicembre 2017
di Corrado De Concini (Autore), Claudio Procesi (Autore)
the case of 2\times 2  matrices is very special and can be described in full detail  see
Rings With Polynomial Identities and Finite Dimensional Representations of Algebras  – 30 dicembre 2020
Eli Aljadeff (Autore), Antonio Giambruno (Autore), Claudio Procesi (Autore), Amitai Regev (Autore)
A: For some Lie groups you will need additional invariants.  For example, for the even orthogonals you will need the  Pfaffian in addition to the Trace.
See for example:
Invariant theory of special orthogonal groups, Helmer Aslaksen, Eng-Chye Tan, Chen-bo Zhu Pacific J. Math. 168(2): 207-215 (1995).
However, you can already see this with one copy of $\mathfrak{so}(2,\mathbb{F})$ and $\mathrm{SO}(2,\mathbb{F})$ acting by conjugation. For in that case the Lie algebra is $\left\{\left(\begin{array}{cc}0&a\\-a&0\end{array}\right)|\ a\in \mathbb{F}\right\}$ and the conjugation action is trivial.  In this case the trace is identically 0 and the Pfaffian is the only useful invariant.
In case you are interested, the Lie group version of this question was addressed by myself and Sikora here:
Varieties of Characters, Sean Lawton & Adam S. Sikora
Algebras and Representation Theory volume 20, pages 1133–1141(2017).
Remark 1: In the above answer, I assumed the OP wanted a fixed representation of $G$.  As noted in the answers to this MO question, if you vary over all representations, then for $k=1$ the answer is yes.  But for $k\geq 2$ the answer appears to me to still be no.  See Theorem 1 in Sikora's paper SO(2n,C)-character varieties are not varieties of characters; it is not exactly the same thing, but it seems to imply the result (see the comments for a strategy to fill in the details).
Remark 2: As noted already by Professor Procesi, his work cited by the OP implies the answer is yes for the Lie groups $\mathrm{GL}_n$, $\mathrm{O}_n$, and $\mathrm{Sp}_{2n}$ (by restricting $k$-tuples of generic matrices to the subvariety $\mathfrak{g}^k$).  From this one can also deduce that the answer is yes for the Lie groups $\mathrm{SL}_n$ and $\mathrm{SO}_{2n+1}$.
Remark 3: As I said in the comments, I believe the work of G. Schwarz probably implies the answer is also yes for a representation of $G_2$ (he also addresses some Spin groups).  The question for the other exceptional groups is open as far as I know.  If I were to guess, I would say it is probably true for all of them except $E_6$ which has additional symmetry as in the case of $\mathrm{SO}_{2n}$ (where the answer is no as I already indicated).
Remark 4: Once one knows the answer for a given $G$, then one also knows it for finite central quotients of $G$ (which is how one goes from the orthogonal case to the special orthogonal case for odd $n$).  Also, if one knows the answer for $G$ and $H$, then one also knows it for $G\times H$.  Putting these observations together with the known cases reduces the entire problem down to the (simply connected forms of the) exceptional groups and the spin groups.
